Question

The point of intersection of the lines represented by equation $2{\left(x+2\right)}^2+3\left(x+2\right)\left(y-2\right)-2{\left(y-2\right)}^2=0$ is

• A. $\left(2,2\right)$
• B. $\left(-2,-2\right)$
• C. $\left(-2,2\right)$
• D. $\left(2,-2\right)$

Answer 1

The correct option is C

$\left(-2,2\right)$

Explanation for the correct answer:

Find the point of intersection of the given lines

Given: $2{\left(x+2\right)}^2+3\left(x+2\right)\left(y-2\right)-2{\left(y-2\right)}^2=0$

Now put, $x+2=X$ and $y-2=Y$

So the equation becomes:

$2X^2+3XY-2Y^2=0$

Which cuts at $X=0$ and $Y=0$

Therefore, we can write

$\Rightarrow x+2=0$ and $y-2=0$

$\Rightarrow x=-2$ and $y=2$

So the point of intersection is $\left(-2,2\right)$.

Hence, option (C) is the answer.